Question

in right triangle abc, if m∠c = 90 and sina = 3/5, cosb is equal to _____. 3/4 4/3 4/5 3/5

Explanation:

Step1: Use the property of right - triangle angles

In a right - triangle \(ABC\) with \(\angle C = 90^{\circ}\), \(\angle A+\angle B = 90^{\circ}\). So, \(\angle B=90^{\circ}-\angle A\).

Step2: Apply the co - function identity

We know the co - function identity \(\cos(90^{\circ}-\theta)=\sin\theta\). Substituting \(\theta = A\) into the identity, we get \(\cos B=\cos(90^{\circ}-A)\).
Since \(\cos(90^{\circ}-A)=\sin A\) and given that \(\sin A=\frac{3}{5}\).

Answer:

\(\frac{3}{5}\)